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I know that the equivalent resistance of this circuit will be

$$R_1-j/(\omega*C_x)+L_1+(R_2-C_1)||L_2$$ $$(3+.814j)-j/(4737*C_x)+[(4j)(30-8j)]/(30-4j)$$

I don't know what to do from here; I could set the top equation equal to Z, but Z is not given. I think I have to use the power factor given, but I have no idea what to do with it. Can someone tell me how power factor comes into all of this?

  • 1
    \$\begingroup\$ The total impedance is \$3\:\Omega+j\:814\:\text{m}\Omega+\left(j\:4\:\Omega\mid\mid\left(30\:\Omega -j\:8\:\Omega\right)\right)+C_x\$. Just work out the first part without \$C_x\$ and express it as a complex number, \$Z\$. If \$\operatorname{Im}\left(Z\right)>0\$, then the addition of a capacitor can achieve the goal. Just solve for \$-j\:\operatorname{Im}\left(Z\right)=\frac{1}{j\:\omega\:C}\$. (You are told \$\omega\$, already.) This is because the imaginary part has to go to zero in this case. \$\endgroup\$ – jonk May 5 '18 at 23:48

A Power factor of unity means that the impedance into the circuit looks purely resistive so, form an equation that describes the input impedance and, equate all terms involving “j” to zero. Solve to find the capacitance and hopefully you have an unambiguous answer.

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